Question: Let $a(x)=-12x^5-7x^3-4x^2+6$, and $b(x)=x$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Answer: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{-12x^5-7x^3-4x^2+6}{x}&=\dfrac{-12 {x^5}-7 {x^3}-4 {x^2}}{ x}+\dfrac{6}{x}\\\\ &={-12x^4-7x^2-4x}+\dfrac{{6}}{x}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${6}$ is less than the degree of $x$, it follows that ${r(x)}={6}$, and ${q(x)}={-12x^4-7x^2-4x}$. To conclude, $q(x)=-12x^4-7x^2-4x$ $r(x)=6$ [Is there another way of doing this?]